Bonsoir, pouvez vous m'aider à faire un exercice de Maths s'il vous plaît ? C pour mardi (Photo en pièce jointe)

Réponse:

oui

Explications étape par étape:

1 . C. f (x) = 1 = 3x²+5x+3

1-3x²+-x-3 = 0

-2-3x²-5x = 0

2+3x²+5x = 0

3x²+2x+3x+2 = 0

3x(x+1)+2(x+1) = 0

(x+1)(3x+2) = 0

x1 = x+1

x1 : -1

et

x2 = 3x+2

x2 : -(2/3)

f (x) = 3 = 3x²+5x+3

0 = 3x²+5x

0-3x²-5x = 0

-x(3x+5) = 0

x(3x+5) = 0

x1 : x = 0

x1 = 0

et

x2 : 3x+5 = 0

x2 = -(5/3)

2. a. f(x) = (2x+5)²-1

4x²+20x+25-1

4x²+20x+24

b. f(x) = (2x+5)²-1

(2x+5-1)(2x+5+1)

(2x+4)(2x+6)

2(x+2)×2(x+3)

2×2(x+2)(x+3)

4(x+2)(x+3)

c. f(x) = 0 = 4x²+20x+24

4x²+20x+24 = 0

x²+5x+6 = 0

x²+3x+2x+6 = 0

x(x+3)+2(x+3) = 0

(x+3)(x+2) = 0

x1 : x+3 = 0

x1 = -3

et

x2 : x+2 = 0

x2 = -2

f(x) = 24 = 4x²+20x+24

0 = 4x²+20x

-4x(x+5) = 0

x(x+5) = 0

x1 : x = 0

x1 = 0

et

x2 : x+5 = 0

x2 = -5

f(x) = 3 = 4x²+20x+24

3-4x²+20x+24 = 0

-21-4x²-20x = 0

4x²+20x+21 = 0

4x²+14x+6x+21 = 0

2x(2x+7)+3(2x+7) = 0

(2x+7)(2x+3) = 0

x1 : 2x+7 = 0

x1 = -(7/2)

et

x2 : 2x+3 = 0

x2 = -(3/2)

3. a. x²-6x+5 = (x-1)(x-5)

= x²-5x-1x+5

= x²-6x+5

b. f(x) > 3

x²-6x+8 > 3

x²-6x+5 > 0

x(x-1)-5(x-1) > 0

(x-1)(x-5) > 0

on résout les inégalités

x-1 > 0 = x > 1

x+5 > 0 = x > 5

et

x-1 < 0 = x < 1

x+5 < 0 = x < 5

Soit

x E ] 5;+infini [

x E ] -infini;1 [

Donc

x E ] -infini;1[ U ] 5;+infini [